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Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle PTQ = 2 \angle OPQ$.
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For correct figure (1/2 Mark)
OP = OQ $\Rightarrow \angle OPQ = \angle OQP$ (1/2 Mark)
$\angle OPQ + \angle OQP + \angle POQ = 180^\circ$ (1/2 Mark)
$\Rightarrow \angle POQ= 180^\circ - 2 \angle OPQ$ -------(i) (1/2 Mark)
Also, $\angle OPT + \angle OQT + \angle POQ + \angle PTQ = 360^\circ$
$\Rightarrow \angle POQ = 180^\circ - \angle PTQ$ -------(ii) (1 Mark)
By (i) and (ii)
$180^\circ - 2 \angle OPQ = 180^\circ - \angle PTQ$
$\Rightarrow 2 \angle OPQ = \angle PTQ$ (1/2 Mark)
OP = OQ $\Rightarrow \angle OPQ = \angle OQP$ (1/2 Mark)
$\angle OPQ + \angle OQP + \angle POQ = 180^\circ$ (1/2 Mark)
$\Rightarrow \angle POQ= 180^\circ - 2 \angle OPQ$ -------(i) (1/2 Mark)
Also, $\angle OPT + \angle OQT + \angle POQ + \angle PTQ = 360^\circ$
$\Rightarrow \angle POQ = 180^\circ - \angle PTQ$ -------(ii) (1 Mark)
By (i) and (ii)
$180^\circ - 2 \angle OPQ = 180^\circ - \angle PTQ$
$\Rightarrow 2 \angle OPQ = \angle PTQ$ (1/2 Mark)